Let $V_n=\operatorname{Var}(S_n)=\sum_{m=1}^nv_m$. Then
$$
\begin{aligned}
\mathbb E[S_{n+1}^2-V_{n+1}\mid\mathcal F_n]
&=S_n^2+v_{n+1}-V_n-v_{n+1}\\
&=S_n^2-V_n.
\end{aligned}
$$
Consequently
$$
\boxed{(S_n^2-\operatorname{Var}(S_n))_{n\geq0}
\text{ is a martingale}.}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 201 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-201/2/c/solution